Direction Angles: Exercises

Problem 1

Find all approximate direction angles of the vector \(\Vect{u} = (4,-2,-4)\) in \(\RNrSpc{3}\).

Solution Answer

Problem 2

Find the angles of the triangle in \(\RNrSpc{3}\) with corners

\[P(1,1,1)\qquad Q(4,3,2)\qquad R(2,3,4)\]

Hint Solution Answer

Solution for Problem 2

The angle of the triangle at \(P\) is the angle between the vectors \(\Vect{x}\) and \(\Vect{y}\) represented by the arrows \(P\) to \(Q\) and \(P\) to \(R\), respectively. We find

\(\Vect{x} = (3,2,1)\) and \(\Vect{y} = (1,2,3)\).

Now the formula \(\DotPr{ \Vect{x} }{ \Vect{y} } = | \Vect{x} | | \Vect{y} | \cos \sphericalangle(\Vect{x} , \Vect{y})\) yields for the angle at \(P\)

\(\cos \sphericalangle(\Vect{x} , \Vect{y} )\)	\(=\)	\(\dfrac{ \DotPr{ \Vect{x} }{ \Vect{y} } }{ \| \Vect{x} \| \| \Vect{y} \|} = \dfrac{10}{14}\)
\(\sphericalangle( \Vect{x} , \Vect{y} )\)	\(\approx\)	\(44.4\quad \text{degree}\)

Similarly, we find that the angle at \(Q\) is approximately \(61.8\) degrees. Consequently, the angle at \(R\) is approximately

\(180\ -\ (44.4 + 61.8)\ =\ 73.8\) degrees.